Answer :
Let's go through the problem step-by-step.
### Part (a) Find the coordinates of the centre of [tex]\( C \)[/tex]
The center of the circle is the midpoint of the diameter [tex]\( PQ \)[/tex]. Given the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- [tex]\( P(-2, 6) \)[/tex]
- [tex]\( Q(4, -1) \)[/tex]
The midpoint formula for points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Applying this formula:
[tex]\[ M_x = \frac{-2 + 4}{2} = \frac{2}{2} = 1 \][/tex]
[tex]\[ M_y = \frac{6 + (-1)}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the coordinates of the centre of circle [tex]\( C \)[/tex] are:
[tex]\[ (1.0, 2.5) \][/tex]
### Part (b) Show that [tex]\( C \)[/tex] has the equation [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]
We start by knowing the standard equation of a circle centered at [tex]\((h, k)\)[/tex] with radius [tex]\( r \)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
For our circle:
[tex]\[ h = 1.0, \quad k = 2.5 \][/tex]
Next, we find the radius [tex]\( r \)[/tex], which is half the length of the diameter [tex]\( PQ \)[/tex].
Using the distance formula to find [tex]\( PQ \)[/tex]:
[tex]\[ PQ = \sqrt{(4 - (-2))^2 + (-1 - 6)^2} \][/tex]
[tex]\[ PQ = \sqrt{6^2 + (-7)^2} \][/tex]
[tex]\[ PQ = \sqrt{36 + 49} \][/tex]
[tex]\[ PQ = \sqrt{85} \][/tex]
Since [tex]\( PQ \)[/tex] is the diameter:
[tex]\[ r = \frac{PQ}{2} = \frac{\sqrt{85}}{2} \][/tex]
So, [tex]\( r^2 = \left(\frac{\sqrt{85}}{2}\right)^2 = \frac{85}{4} \)[/tex].
Now, substitute [tex]\( h, k, \)[/tex] and [tex]\( r^2 \)[/tex] into the standard circle equation and simplify:
[tex]\[ (x - 1)^2 + (y - 2.5)^2 = \frac{85}{4} \][/tex]
[tex]\[ x^2 - 2x + 1 + y^2 - 5y + 6.25 = \frac{85}{4} \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 = \frac{85}{4} \][/tex]
Converting [tex]\( \frac{85}{4} \)[/tex] to a decimal:
[tex]\[ \frac{85}{4} = 21.25 \][/tex]
Thus:
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 = 21.25 \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 - 21.25 = 0 \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y - 14 = 0 \][/tex]
This confirms the equation of circle [tex]\( C \)[/tex].
### Part (c) Show that [tex]\( R \)[/tex] lies on [tex]\( C \)[/tex] and hence, state the size of [tex]\( \angle PRQ \)[/tex] in degrees
First, verify that [tex]\( R(2,7) \)[/tex] satisfies the circle's equation [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]:
Substitute [tex]\( (2, 7) \)[/tex] into the equation:
[tex]\[ x^2 + y^2 - 2x - 5y - 14 = 0 \][/tex]
[tex]\[ 2^2 + 7^2 - 2(2) - 5(7) - 14 = 0 \][/tex]
[tex]\[ 4 + 49 - 4 - 35 - 14 = 0 \][/tex]
[tex]\[ 53 - 53 = 0 \][/tex]
[tex]\[ 0 = 0 \][/tex]
As this is true, point [tex]\( R \)[/tex] lies on circle [tex]\( C \)[/tex].
To find [tex]\( \angle PRQ \)[/tex], note that since the points lie on the circle and since [tex]\( PQ \)[/tex] is a diameter, [tex]\( \angle PRQ \)[/tex] is a right angle. This is because any triangle inscribed in a circle where one side is the diameter is a right triangle (Thales' theorem).
Thus, the size of [tex]\( \angle PRQ \)[/tex] is:
[tex]\[ 90 \text{ degrees} \][/tex]
So, the answers are:
- The coordinates of the centre of [tex]\( C \)[/tex] are [tex]\( (1.0, 2.5) \)[/tex]
- The equation of [tex]\( C \)[/tex] is verified as [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]
- Point [tex]\( R \)[/tex] lies on [tex]\( C \)[/tex] and [tex]\( \angle PRQ \)[/tex] is [tex]\( 90 \)[/tex] degrees.
### Part (a) Find the coordinates of the centre of [tex]\( C \)[/tex]
The center of the circle is the midpoint of the diameter [tex]\( PQ \)[/tex]. Given the coordinates of [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- [tex]\( P(-2, 6) \)[/tex]
- [tex]\( Q(4, -1) \)[/tex]
The midpoint formula for points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Applying this formula:
[tex]\[ M_x = \frac{-2 + 4}{2} = \frac{2}{2} = 1 \][/tex]
[tex]\[ M_y = \frac{6 + (-1)}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the coordinates of the centre of circle [tex]\( C \)[/tex] are:
[tex]\[ (1.0, 2.5) \][/tex]
### Part (b) Show that [tex]\( C \)[/tex] has the equation [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]
We start by knowing the standard equation of a circle centered at [tex]\((h, k)\)[/tex] with radius [tex]\( r \)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
For our circle:
[tex]\[ h = 1.0, \quad k = 2.5 \][/tex]
Next, we find the radius [tex]\( r \)[/tex], which is half the length of the diameter [tex]\( PQ \)[/tex].
Using the distance formula to find [tex]\( PQ \)[/tex]:
[tex]\[ PQ = \sqrt{(4 - (-2))^2 + (-1 - 6)^2} \][/tex]
[tex]\[ PQ = \sqrt{6^2 + (-7)^2} \][/tex]
[tex]\[ PQ = \sqrt{36 + 49} \][/tex]
[tex]\[ PQ = \sqrt{85} \][/tex]
Since [tex]\( PQ \)[/tex] is the diameter:
[tex]\[ r = \frac{PQ}{2} = \frac{\sqrt{85}}{2} \][/tex]
So, [tex]\( r^2 = \left(\frac{\sqrt{85}}{2}\right)^2 = \frac{85}{4} \)[/tex].
Now, substitute [tex]\( h, k, \)[/tex] and [tex]\( r^2 \)[/tex] into the standard circle equation and simplify:
[tex]\[ (x - 1)^2 + (y - 2.5)^2 = \frac{85}{4} \][/tex]
[tex]\[ x^2 - 2x + 1 + y^2 - 5y + 6.25 = \frac{85}{4} \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 = \frac{85}{4} \][/tex]
Converting [tex]\( \frac{85}{4} \)[/tex] to a decimal:
[tex]\[ \frac{85}{4} = 21.25 \][/tex]
Thus:
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 = 21.25 \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y + 7.25 - 21.25 = 0 \][/tex]
[tex]\[ x^2 + y^2 - 2x - 5y - 14 = 0 \][/tex]
This confirms the equation of circle [tex]\( C \)[/tex].
### Part (c) Show that [tex]\( R \)[/tex] lies on [tex]\( C \)[/tex] and hence, state the size of [tex]\( \angle PRQ \)[/tex] in degrees
First, verify that [tex]\( R(2,7) \)[/tex] satisfies the circle's equation [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]:
Substitute [tex]\( (2, 7) \)[/tex] into the equation:
[tex]\[ x^2 + y^2 - 2x - 5y - 14 = 0 \][/tex]
[tex]\[ 2^2 + 7^2 - 2(2) - 5(7) - 14 = 0 \][/tex]
[tex]\[ 4 + 49 - 4 - 35 - 14 = 0 \][/tex]
[tex]\[ 53 - 53 = 0 \][/tex]
[tex]\[ 0 = 0 \][/tex]
As this is true, point [tex]\( R \)[/tex] lies on circle [tex]\( C \)[/tex].
To find [tex]\( \angle PRQ \)[/tex], note that since the points lie on the circle and since [tex]\( PQ \)[/tex] is a diameter, [tex]\( \angle PRQ \)[/tex] is a right angle. This is because any triangle inscribed in a circle where one side is the diameter is a right triangle (Thales' theorem).
Thus, the size of [tex]\( \angle PRQ \)[/tex] is:
[tex]\[ 90 \text{ degrees} \][/tex]
So, the answers are:
- The coordinates of the centre of [tex]\( C \)[/tex] are [tex]\( (1.0, 2.5) \)[/tex]
- The equation of [tex]\( C \)[/tex] is verified as [tex]\( x^2 + y^2 - 2x - 5y - 14 = 0 \)[/tex]
- Point [tex]\( R \)[/tex] lies on [tex]\( C \)[/tex] and [tex]\( \angle PRQ \)[/tex] is [tex]\( 90 \)[/tex] degrees.
